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3 years ago

3(2x-4)>-12 solve for x?

Mathematics

1 answer:

Darina [25.2K]3 years ago

7 0

Answer:

x > 0

Step-by-step explanation:

6x -12 > -12

6x > -12 +12

6x > 0

x > 0/6

x > 0

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1^2 + 2^2 + 3^2 +.... + 28^2 + 29^2 + 30^2 = ......<br> Find the value <br> Thank you!

liberstina [14]

Step-by-step explanation:

The sum of consecutive squares is

$1^2+2^2+\dots+n^2=\frac16n(n+1)(2n+1)$

Therefore

$1^2+2^2+\dots+30^2=\frac16(30)(31+1)(2\cdot30+1) = 9455$

8 0

3 years ago

Which of the following number lines represents the solution set of the inequality -1/3 x + 4 < 3?

natima [27]

Solve for x.
x > 3
(It goes to the right, with the starting point open)

8 0

3 years ago

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alex41 [277]

The kite is 1500 feet above the ground

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4 years ago

A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some ty

Sophie [7]

Answer:

The interval is $0.7187 < p < 2.421$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 100$

The population proportion is $p = 0.81$

The confidence level is C = 98%

The level of significance is mathematically evaluated as

$\alpha = 100 -98$

$\alpha = 2%$ %

$\alpha = 0.02$

Here this level of significance represented the left and the right tail

The degree of freedom is evaluated as

$df = n-1$

substituting value

$df = 100 - 1$

$df = 99$

Since we require the critical value of one tail in order to evaluate the 98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The critical value of $\frac{\alpha}{2}$ and the evaluated degree of freedom is

$t_{df , \alpha } = t_{99 , \frac{0.02}{2} } = 2.33$

this is obtained from the critical value table

The standard error is mathematically evaluated as

$SE = \sqrt{\frac{p(1-p )}{n} }$

substituting value

$SE = \sqrt{\frac{0.81(1-0.81 )}{100} }$

$SE = 0.0392$

The 98% confidence interval is evaluated as

$p - t_{df , \frac{\alpha }{2} } * SE < p < p + t_{df , \frac{\alpha }{2} }$

substituting value

$0.81 - 2.33 * 0.0392 < p < 0.81 +2.33 * 0.0392$

$0.7187 < p < 2.421$

4 0

3 years ago

What is 65(k+1)/9 simplified

nadezda [96]

Answer:

65k/9+65/9

Step-by-step explanation:

65(K+1)/9

you would distribute the 65 so that would be

65k+65

65k/9+65/9.

This is the max you can solve If you have the number that this equation equal to, them you would go and further solve for k.

3 0

3 years ago